A and B are positive acute angles satisfying the equations
`3cos^(2)A+2cos^(2)B=4and (3sinA)/(sinB)=(2cosB)/(cosA), then A+2B` is equal to

Options

A. `pi//4`
B. `pi//3`
C. `pi//6`
D. `pi//2`

Answer

D

Explanation

From the given relation, we have
`2 cos^(2) B-1=3(1-cos^(2)A)`
or `cos 2B = 3 sin^(2)A` …(1)
`(3 sin A)/(sin B)=(2 cos B)/(cos A)`
`rArr (3)/(2) sin 2A = sin 2B`
`rArr 3 sin 2A = 2 sin 2B` …(2)
Now cos (A + 2B)
`= cos A cos 2B - sin A sin 2B`
`= cos A(3 sin^(2)A)-sin A.(3)/(2)sin 2A`
`= 3 cos A sin^(2)A-3 sin^(2)A cos A = 0`
`therefore A+2B=90^(@)`